killjoy1990xb9

2022-01-20

Solving simple differential equation

$(t+4)dx=4(1+{x}^{2})dt$

Bob Huerta

Beginner2022-01-20Added 41 answers

As you rearranged it in your post, we have

$\int \frac{dx}{1+{x}^{2}}=4\int \frac{dt}{t+4}$

As you discovered, the LHS is arctan. To find the value of the RHS, we may say$t+4=u\text{}\text{so}\text{}du=dt$ . Thus the RHS can be rewritten as

$4\int \frac{du}{u}=4\mathrm{log}\left|u\right|+{C}_{2}$

Because we want this integral to be in respect to t, we have$u=t+4$ , so

$\mathrm{arctan}x+{C}_{1}=4\mathrm{log}|t+4|+{C}_{2}$

To confirm that the integral we solved is indeed correct, you can simply differentiate$4\mathrm{log}|t+4|+{C}_{2}$

Thus$x=\mathrm{tan}(4\mathrm{log}|t+4|+{C}_{2}-{C}_{1})$

and$t=\mathrm{exp}\left(\frac{1}{4}(\mathrm{arctan}x+{C}_{1}-{C}_{2})\right)-4$

As you discovered, the LHS is arctan. To find the value of the RHS, we may say

Because we want this integral to be in respect to t, we have

To confirm that the integral we solved is indeed correct, you can simply differentiate

Thus

and

Dawn Neal

Beginner2022-01-21Added 35 answers

Here are some proceeding steps:

- Seperating the variables you get

$\frac{dx}{1+{x}^{2}}=4\cdot \frac{dt}{t+4}$

- Integrating both sides you have:

$\mathrm{tan}}^{-1}\left(x\right)+{C}_{1}=4\cdot \mathrm{log}|t+4|+{C}_{2$ .

- Seperating the variables you get

- Integrating both sides you have: